Ajay S.

asked • 08/11/19

Physics Question

Assuming that the earth re-emits all the radiations it receives from the sum, determine its temperature, (For earth, E = 1.39 × 103 w/m2 , Stefan’s constant. = σ = 5.735 × 10-8 w/m2 k4 )

Steven W.

tutor
The answer depends on what is meant by "the Earth." Is it the ground? Everything up to the troposphere? The "top-of-atmosphere" (usually taken as the top of the mesosphere)? Everything up to "deep space"? The reason it matters is that the temperature of the environment needs to be considered. The "environment" above the ground is at a much different temperature than at the edge of deep space, and there is great variation between. More specificity would be needed. The formula in question is the Stefan-Boltzmann law: P = e(sigma)A(T^4 - T_env^4) where P =emitted power e = emissivity (assumed to be 1 for a blackbody, which we are calling the Earth in this case... since we are told to "assum[e] that the [E]arth re-emits all the radiation...it receives"; which is the definition of a blackbody) sigma = Stefan-Botlzmann constant, listed in your text A = surface area of the body in question T = temperature of the body (in kelvin, if you use the value of (sigma) you listed) T_env = temperature of the environment (in kelvin) You are given E = power flux (power emitted per unit area) = P/A So the above expression can be rewritten: E = (sigma)(T^4 - T_env^4) (assuming e = 1, as stated above) So T_env is needed. For deep space, it would be about 2.75 K. Right above the ground, it would be quite different, as it would be at different levels of the atmosphere.
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08/12/19

1 Expert Answer

By:

Tom N. answered • 08/13/19

Tutor
4.9 (503)

MS in Geophysics.

Steven W.

tutor
This does assume there is no significant return radiation from the surroundings, which is probably reasonable if the surroundings are deep space (~3 K), though less so if we are talking about the Earth as ground and the atmosphere right above it
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08/13/19

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